$\overline{AB}$ = $2\sqrt{41}$ $\overline{BC} = {?}$ $A$ $C$ $B$ $2\sqrt{41}$ $?$ $ \sin( \angle BAC ) = \frac{4\sqrt{41} }{41}, \cos( \angle BAC ) = \frac{5\sqrt{41} }{41}, \tan( \angle BAC ) = \dfrac{4}{5}$
Explanation: $\overline{AB}$ is the hypotenuse $\overline{BC}$ is opposite to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle BAC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}}= \frac{\overline{BC}}{2\sqrt{41}} $ $ \overline{BC}=2\sqrt{41} \cdot \sin( \angle BAC ) = 2\sqrt{41} \cdot \frac{4\sqrt{41} }{41} = 8$